Proof of Euclid's Lemma without using Bezout's identity or Induction

Question

I'm investigating Euclid's Lemma and the standard proofs I've come across invoke Bezout's identity or use induction. There is nothing wrong with both of these methods, but I was seeking a proof that was more elementary as I'm just starting out with proof-based math. I saw a few proofs posted on this site but I'm struggling to understand those, so I thought I'd just give it a go myself.

Euclid's Lemma: If a prime $p$ divides the product ab of two integers $a$ and $b$, then $p$ must divide at least one of those integers $a$ or $b$.

I state the Lemma as:

$Prime(p) \land (p \mid ab) \Rightarrow (p \mid a) \lor (p \mid b) \text{ for any integers } a,b$

If $p | ab$, then by definition $ab$ can be expressed as $ab = qp$ where $q$ is an integer. Restated in terms of $q$: $q = ab/p$.

Let's assume $p$ does not divide $a$. Then $a$ and $p$ are coprime because $p$ is prime. In other words, $a$ does not divide $p$.

Then let's investigate the possibilities for the value of $b$ for $q$ to be an integer:

$q = (a/p)\cdot b$

$a/p$ is expressed as a fraction in lowest terms because $p$ only divides itself and we assumed $p$ does not divide $a$ earlier. We need to multiple this fraction by some integer $b$ so that the product is no longer a fraction.

There are two possibilities here such that $q$ is an integer: 1) $b$ is some integer multiple of $p$, or 2) $ab = p$ or $-p$, rendering $(a\cdot b)/p = 1$ or $-1$. However, it cannot be the case that $ab = p$ because that would imply $p$ is composite (not prime). So it must be that $b$ is some integer multiple of $p$, or in other words, $p$ divides $b$. This clearly applies vice versa as well; assuming $p$ does not divide $b$ implies $p$ divides $a$.

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I sense I've made an error here as I don't think the proof could be so simple, as that would obviate the need for more elaborate proofs. Thoughts or corrections?

Answer 1

The proof starts to fall apart at the line "There are two possibilities here...".

At that point, we know the following things:

1. There is no integer dividing both $p$ and $a$.
2. $(a/p) \cdot b$ is an integer $q$.
3. And of course, $p$ is prime

There is a third possibility that $a \cdot b$ is not equal to $p$, $p$ is not a factor of either $a$ or $b$, and yet $p$ is a factor of $a \cdot b$. So why are you excluding this possibility?

1. If you are not using the assumption that $p$ is prime to justify this step, then you are wrong. This third possibility occurs, for instance, if $p=6$, $a=4$, $b=3$.
2. If you are using the assumption that $p$ is prime to justify this step, then you're implicitly using the very thing you're trying to prove as a hypothesis, which is the biggest thing you can never do in a proof.

The reason that your proof won't work without Bezout's Lemma or induction is somewhat deep. In higher mathematics, we study objects called rings which satisfy all the usual laws of integer arithmetic. In some rings, Euclid's Lemma fails to hold even though all of the rules of arithmetic are true.

And so to prove Euclid's Lemma for the integers, you need some special property of the integers. The power of induction is one special property, Bezout's Lemma is another.